Explain how to solve the following IVP.
\[ -12 \, {y} + 12 \, \delta\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -8 \]
Hint: \( \frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}} \).
Answer:
\[ \mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 4} - \frac{8}{s^{2} - 4} \]
\[ \mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s + 2} - \frac{e^{\left(-s\right)}}{s - 2} + \frac{2}{s + 2} - \frac{2}{s - 2} \]
\[ {y} = -e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{\left(-2 \, t\right)} \]